MATHS
QUERY

Perimeter is the length of the boundary around any closed figure.

The boundary of a triangle consists of its three sides. Therefore, sum of length of three sides of triangle is
called its perimeter.

Units of perimeter measurement of a triangle are taken the same as units of length of sides of the triangle.

The few examples of units of length on the side of a triangle are meter (m), centimeter (cm) etc..
So, the units of perimeter are also taken as meter (m), centimeter (cm) etc..

Perimeter of triangle ΔABC

In this ΔABC, we have

length of AB = a

length of BC = b

and length of CA = c

So, perimeter of ΔABC is sum of length of sides AB, BC and CA
**i.e. perimeter of ΔABC** = a + b + c

Equilateral triangle ΔABC with sides length a

In an equilateral triangle, where all sides are of equal length.

i.e. AB = a, BC = a and CA = a

So, **perimeter of equilateral ΔABC** = a + a + a = 3a

Isosceles triangle ΔABC with sides length a, b and a

In isosceles triangle, where any two sides are of equal length.

i.e. AB = a, BC = a and CA = b

So, **perimeter of isosceles ΔABC** = a + a + b = 2a + b

Example

**Find perimeter of ΔABC with length of its sides as given below:**

AB=2cm, BC=4cm, CA=6cm

So, perimeter of ΔABC = AB + BC + CA

= 2 + 4 + 6

= 12cm

Area is the total amount of space occupied by any closed figure.

Area is measured in square units.

The few examples of units of area of triangle are meter^{2} or m^{2} read as meter square,
centimeter^{2} or cm^{2} read as centimeter square etc.

Area of any triangle = $\frac{1}{2}$ × base × height

Let’s understand it from following ΔABC

Area of triangle ΔABC

where, BC is base, length of BC = b

and AO is height, length of AO = h

Therefore, area of ΔABC = $\frac{1}{2}$ × b × h

Area of equilateral triangle ΔABC

In above equilateral ΔABC, where all sides of triangle are equal in length, its area is calculated
as:

area of ΔABC = $\frac{$ \sqrt{3}$}{4}$
a^{2}
**Let’s see how it is calculated?**

Here, in right angle ΔAOC

a^{2} = h^{2} + ($\frac{\mathrm{a}}{2}$)^{2}
by
Pythagoras theorem

a^{2} = h^{2} +
$\frac{\mathrm{$ {a}^{2}$}}{4}$

a^{2} –
$\frac{\mathrm{$ {a}^{2}$}}{4}$
= h^{2}

$\frac{\mathrm{$ {\mathrm{4a}}^{2}\u2013{a}^{2}$}}{4}$
= h^{2}

$\frac{\mathrm{$ {\mathrm{3a}}^{2}$}}{4}$ = h^{2}

$\frac{$ \sqrt{3}$}{4}$ a = h

Area of Δ =
$\frac{1}{2}$
× a ×
$\frac{$ \sqrt{3}$}{2}$ a

Area of Δ = $\frac{$ \sqrt{3}$}{4}$
a^{2}

Area of isosceles triangle ΔABC

area of isosceles ΔABC =
$\frac{\text{b}\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{4}$
**Let’s see how it is calculated?**

Here, in right angle ΔAOC

(a)^{2} = (h)^{2} + ($\frac{\mathrm{b}}{2}$)^{2}
by
Pythagoras theorem

a^{2} = h^{2} +
$\frac{\mathrm{$ {b}^{2}$}}{4}$

a^{2} – $\frac{\mathrm{$ {b}^{2}$}}{4}$ = h^{2}

$\frac{\mathrm{$ {\mathrm{4a}}^{2}\u2013{b}^{2}$}}{4}$
= h^{2}

$\sqrt{\frac{{\mathrm{4a}}^{2}\u2013{b}^{2}}{4}}$ = h

$\frac{\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{2}$ = h

we know, area of Δ = $\frac{1}{2}$ × b × h

∴ area of ΔABC = $\frac{1}{2}$ × b ×
$\frac{\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{2}$

or area of ΔABC =
$\frac{\text{b}\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{4}$